Problem: $ C = \left[\begin{array}{rrr}1 & 3 & 1 \\ 2 & 2 & 4\end{array}\right]$ $ A = \left[\begin{array}{rr}5 & -1 \\ 1 & -1 \\ 4 & 5\end{array}\right]$ What is $ C A$ ?
Solution: Because $ C$ has dimensions $(2\times3)$ and $ A$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ C A = \left[\begin{array}{rrr}{1} & {3} & {1} \\ {2} & {2} & {4}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{-1} \\ {1} & \color{#DF0030}{-1} \\ {4} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ C$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ C$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ C$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{5}+{3}\cdot{1}+{1}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ C$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{5}+{3}\cdot{1}+{1}\cdot{4} & ? \\ {2}\cdot{5}+{2}\cdot{1}+{4}\cdot{4} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ C$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{5}+{3}\cdot{1}+{1}\cdot{4} & {1}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{5} \\ {2}\cdot{5}+{2}\cdot{1}+{4}\cdot{4} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{5}+{3}\cdot{1}+{1}\cdot{4} & {1}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{-1}+{1}\cdot\color{#DF0030}{5} \\ {2}\cdot{5}+{2}\cdot{1}+{4}\cdot{4} & {2}\cdot\color{#DF0030}{-1}+{2}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}12 & 1 \\ 28 & 16\end{array}\right] $